For the integral of (x − 4y) over the rectangle −1 ≤ x ≤ 2, 0 ≤ y ≤ 1: describe the region, evaluate the double integral, then reverse the order of integration and evaluate again.

Step-by-step solution

1. 1

   (a) Read the limits. The inner variable is $x$ with constant limits $-1$ and $2$; the outer is $y$ from $0$ to $1$. Because all four limits are constants, the region is a rectangle:

   $D=\{(x,y):-1\le x\le 2,\ 0\le y\le 1\}$

   with corners $(-1,0)$, $(2,0)$, $(2,1)$ and $(-1,1)$.

2. 2

   (b) Integrate with respect to x, treating y as constant.

   $\int_{-1}^{2}(x-4y)\,dx=\left[\frac{x^{2}}{2}-4yx\right]_{-1}^{2}=(2-8y)-\left(\tfrac12+4y\right)=\frac{3}{2}-12y$

3. 3

   Integrate the result over y.

   $\int_{0}^{1}\left(\frac32-12y\right)dy=\left[\frac{3y}{2}-6y^{2}\right]_{0}^{1}=\frac32-6=-\frac92$

4. 4

   (c) Reverse the order. Over a rectangle the limits simply swap places — no case-splitting is needed, because neither variable's range depends on the other:

   $\int_{-1}^{2}\!\!\int_{0}^{1}(x-4y)\,dy\,dx$

5. 5

   Evaluate in the new order. Inner integral over $y$:

   $\int_{0}^{1}(x-4y)\,dy=\left[xy-2y^{2}\right]_{0}^{1}=x-2$

   then over $x$:

   $\int_{-1}^{2}(x-2)\,dx=\left[\frac{x^{2}}{2}-2x\right]_{-1}^{2}=(2-4)-\left(\tfrac12+2\right)=-2-\tfrac52=-\frac92$

   $\boxed{-\dfrac92}$

6. 6

   Confirm with a shortcut. Because the integrand splits and the region is a rectangle, $\iint_D(x-4y)\,dA=\left(\int_{-1}^{2}x\,dx\right)(1)-4\left(\int_{0}^{1}y\,dy\right)(3)=\tfrac32-4\left(\tfrac12\right)(3)=\tfrac32-6=-\tfrac92$, matching both orders — an instance of Fubini's theorem, which guarantees the two orders agree for a continuous integrand on a rectangle.